Python 读取文件数据而不将其保存在 Flask 中
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Read file data without saving it in Flask
提问by user2480542
I am writing my first flask application. I am dealing with file uploads, and basically what I want is to read the data/content of the uploaded file without saving it and then print it on the resulting page. Yes, I am assuming that the user uploads a text file always.
我正在编写我的第一个烧瓶应用程序。我正在处理文件上传，基本上我想要的是在不保存的情况下读取上传文件的数据/内容，然后将其打印在结果页面上。是的，我假设用户总是上传一个文本文件。
Here is the simple upload function i am using:
这是我正在使用的简单上传功能：
@app.route('/upload/', methods=['GET', 'POST'])
def upload():
if request.method == 'POST':
file = request.files['file']
if file:
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
a = 'file uploaded'
return render_template('upload.html', data = a)
Right now, I am saving the file, but what I need is that 'a' variable to contain the content/data of the file .. any ideas?
现在，我正在保存文件，但我需要的是“a”变量来包含文件的内容/数据......有什么想法吗？
采纳答案by tbicr
FileStoragecontains streamfield. This object must extend IO or file object, so it must contain readand other similar methods. FileStoragealso extend streamfield object attributes, so you can just use file.read()instead file.stream.read(). Also you can use saveargument with dstparameter as StringIOor other IO or file object to copy FileStorage.streamto another IO or file object.
FileStorage包含stream字段。这个对象必须扩展IO或文件对象，所以它必须包含read和其他类似的方法。FileStorage还扩展stream字段对象属性，因此您可以file.read()改为使用file.stream.read(). 您也可以使用save带dst参数的参数 asStringIO或其他 IO 或文件对象来复制FileStorage.stream到另一个 IO 或文件对象。
See documentation: http://flask.pocoo.org/docs/api/#flask.Request.filesand http://werkzeug.pocoo.org/docs/datastructures/#werkzeug.datastructures.FileStorage.
请参阅文档：http: //flask.pocoo.org/docs/api/#flask.Request.files和http://werkzeug.pocoo.org/docs/datastructures/#werkzeug.datastructures.FileStorage。
回答by Dimitry Miles
If you want to use standard Flask stuff - there's no way to avoid saving a temporary file if the uploaded file size is > 500kb. If it's smaller than 500kb - it will use "BytesIO", which stores the file content in memory, and if it's more than 500kb - it stores the contents in TemporaryFile() (as stated in the werkzeug documentation). In both cases your script will block until the entirety of uploaded file is received.
如果您想使用标准的 Flask 内容 - 如果上传的文件大小 > 500kb，则无法避免保存临时文件。如果它小于 500kb - 它将使用“BytesIO”，它将文件内容存储在内存中，如果它超过 500kb - 它会将内容存储在 TemporaryFile() 中（如werkzeug 文档中所述）。在这两种情况下，您的脚本都会阻塞，直到收到整个上传的文件。
The easiest way to work around this that I have found is:
我发现解决此问题的最简单方法是：
1) Create your own file-like IO class where you do all the processing of the incoming data
1) 创建您自己的类似文件的 IO 类，您可以在其中对传入数据进行所有处理
2) In your script, override Request class with your own:
2）在你的脚本中，用你自己的覆盖请求类：
class MyRequest( Request ):
def _get_file_stream( self, total_content_length, content_type, filename=None, content_length=None ):
return MyAwesomeIO( filename, 'w' )
3) Replace Flask's request_class with your own:
3) 用你自己的替换 Flask 的 request_class：
app.request_class = MyRequest
4) Go have some beer :)
4) 去喝点啤酒 :)
回答by lalit
In case we want to dump the in memory file to disk. This code can be used
如果我们想将内存文件转储到磁盘。可以使用此代码
if isinstanceof(obj,SpooledTemporaryFile):
obj.rollover()
回答by Deepak Sharma
We simply did:
我们只是做了：
import io
from pathlib import Path
def test_my_upload(self, accept_json):
"""Test my uploads endpoint for POST."""
data = {
"filePath[]": "/tmp/bin",
"manifest[]": (io.StringIO(str(Path(__file__).parent /
"path_to_file/npmlist.json")).read(),
'npmlist.json'),
}
headers = {
'a': 'A',
'b': 'B'
}
res = self.client.post(api_route_for('/test'),
data=data,
content_type='multipart/form-data',
headers=headers,
)
assert res.status_code == 200
回答by Tri?t Nguy?n V?nh
in function
在功能上
def handleUpload():
if 'photo' in request.files:
photo = request.files['photo']
if photo.filename != '':
image = request.files['photo']
image_string = base64.b64encode(image.read())
image_string = image_string.decode('utf-8')
#use this to remove b'...' to get raw string
return render_template('handleUpload.html',filestring = image_string)
return render_template('upload.html')
in html file
在 html 文件中
<html>
<head>
<title>Simple file upload using Python Flask</title>
</head>
<body>
{% if filestring %}
<h1>Raw image:</h1>
<h1>{{filestring}}</h1>
<img src="data:image/png;base64, {{filestring}}" alt="alternate" />.
{% else %}
<h1></h1>
{% endif %}
</body>
回答by TGanoe
I was trying to do the exact same thing, open a text file (a CSV for Pandas actually). Don't want to make a copy of it, just want to open it. The form-WTF has a nice file browser, but then it opens the file and makes a temporary file, which it presents as a memory stream. With a little work under the hood,
我试图做完全相同的事情，打开一个文本文件（实际上是 Pandas 的 CSV）。不想复制它，只想打开它。form-WTF 有一个很好的文件浏览器，但是它会打开文件并创建一个临时文件，它以内存流的形式呈现。在引擎盖下做一点工作，
form = UploadForm()
if form.validate_on_submit():
filename = secure_filename(form.fileContents.data.filename)
filestream = form.fileContents.data
filestream.seek(0)
ef = pd.read_csv( filestream )
sr = pd.DataFrame(ef)
return render_template('dataframe.html',tables=[sr.to_html(justify='center, classes='table table-bordered table-hover')],titles = [filename], form=form)
回答by jamartincelis
I share my solution (assuming everything is already configured to connect to google bucket in flask)
我分享我的解决方案（假设一切都已配置为连接到烧瓶中的谷歌存储桶）
from google.cloud import storage
@app.route('/upload/', methods=['POST'])
def upload():
if request.method == 'POST':
# FileStorage object wrapper
file = request.files["file"]
if file:
os.environ["GOOGLE_APPLICATION_CREDENTIALS"] = app.config['GOOGLE_APPLICATION_CREDENTIALS']
bucket_name = "bucket_name"
storage_client = storage.Client()
bucket = storage_client.bucket(bucket_name)
# Upload file to Google Bucket
blob = bucket.blob(file.filename)
blob.upload_from_string(file.read())
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